Let $a(x)=-14x^7+8x^5-6x^4+5x$, and $b(x)=x^3$. When dividing $a$ by $b$, we can find the unique quotient polynomial $q$ and remainder polynomial $r$ that satisfy the following equation: $\dfrac{a(x)}{b(x)}=q(x) + \dfrac{r(x)}{b(x)}$, where the degree of $r(x)$ is less than the degree of $b(x)$. What is the quotient, $q(x)$ ? $ q(x)=$ What is the remainder, $r(x)$ ? $r(x)=$
Answer: Note that $a(x)$ has a higher degree than $b(x)$. This allows us to find a non-zero quotient polynomial, $q(x)$. [Why is this important?] Let's rewrite the fraction to cancel common factors: $ \begin{aligned} \dfrac{a(x)}{b(x)}=\dfrac{-14x^7+8x^5-6x^4+5x}{x^3}&=\dfrac{-14 {x^7}+8 {x^5}-6 {x^4}}{ {x^3}}+\dfrac{5x}{x^3}\\\\ &={-14x^4+8x^2-6x}+\dfrac{{5x}}{x^3}\\\\ &={q(x)} + \dfrac{{r(x)}}{b(x)}\end{aligned}$ Since the degree of ${5x}$ is less than the degree of $x^3$, it follows that ${r(x)}={5x}$, and ${q(x)}={-14x^4+8x^2-6x}$. To conclude, $q(x)=-14x^4+8x^2-6x$ $r(x)=5x$ [Is there another way of doing this?]